# What is the equation of the line normal to  f(x)=(x-4)(x-2)+2x+4 at  x=-2?

Jan 14, 2016

$y - 24 = \frac{1}{8} \left(x + 2\right)$

#### Explanation:

First, simplify $f \left(x\right)$.

$f \left(x\right) = {x}^{2} - 4 x - 2 x + 8 + 2 x + 4$
$f \left(x\right) = {x}^{2} - 4 x + 12$

Find the point the tangent line will intercept.

$f \left(- 2\right) = {\left(- 2\right)}^{2} - 4 \left(- 2\right) + 12 = 24$

The tangent line passes through the point $\left(- 2 , 24\right)$.

To find the slope of the tangent line, find $f ' \left(- 2\right)$. First, find $f ' \left(x\right)$ using the power rule.

$f ' \left(x\right) = 2 x - 4$

The slope of the tangent line is:

$f ' \left(- 2\right) = 2 \left(- 2\right) - 4 = - 8$

However, the question asks for the normal line. The normal line is simply the line perpendicular to the tangent line at the point on the function.

Since perpendicular lines have opposite reciprocal slopes and the slope of the tangent line is $- 8$, the slope of the (perpendicular) normal line is $\frac{1}{8}$.

We know the slope is $\frac{1}{8}$ and the line passes through the point $\left(- 2 , 24\right)$. Relate this information in an equation in point-slope form.

$y - 24 = \frac{1}{8} \left(x + 2\right)$

Check a graph:

graph{((x-4)(x-2)+2x+4-y)(y-24.125(x+2))=0 [-15, 20, -29.3, 200]}